There is given a convex quadrilateral
![ABCD](/media/m/9/c/e/9ce25711ba18d9663b73c3580de4bf5a.png)
. Prove that there exists a point
![P](/media/m/9/6/8/968d210d037e7e95372de185e8fb8759.png)
inside the quadrilateral such that
if and only if the diagonals
![AC](/media/m/6/4/7/647ef3a5d68f07d59d84afe03a9dc655.png)
and
![BD](/media/m/1/1/f/11f65a804e5c922ee28a53b1df04d138.png)
are perpendicular.
Proposed by Dukan Dukic, Serbia
%V0
There is given a convex quadrilateral $ABCD$. Prove that there exists a point $P$ inside the quadrilateral such that
$$\angle PAB + \angle PDC = \angle PBC + \angle PAD = \angle PCD + \angle PBA = \angle PDA + \angle PCB = 90^{\circ}$$
if and only if the diagonals $AC$ and $BD$ are perpendicular.
Proposed by Dukan Dukic, Serbia